LearnYMA01WME01 · Mechanics 1Dynamics of a particle in a straight lineFriction and inclined planes

Dynamics of a particle in a straight line · Subtopic 2 of 4

Friction and Inclined Planes

Friction opposes motion (or attempted motion) and can grow up to $\mu R$ . Once sliding, it equals $\mu R$ . On a slope, resolve along and perpendicular to the plane: the weight gives $mg\sin\theta$ down the slope and $mg\cos\theta$ into it, so $R=mg\cos\theta$ when nothing else pushes.

Edexcel YMA01 WME01≈ 40 minRevision 5 · updated 2026-10-27

Section 1 of 14

Learning objectives

Use $F\le\mu R$ , and $F=\mu R$ when sliding.
Find the normal reaction, including with a force at an angle.
Resolve parallel and perpendicular to an inclined plane.
Find accelerations and coefficients of friction.

Pause and think

You push a heavy sofa gently and it doesn't move. Is there friction? How big is it?

Section 2 of 14

Core idea

Start hereWhy steep slopes are slippery

A box rests on a gentle ramp but slides on a steep one. Two things change as the ramp tilts: the pull down the slope grows, and the normal reaction (which sets the maximum friction) shrinks.

The big idea

$F\le\mu R$ , along the surface, opposing motion; $F=\mu R$ when sliding. On a slope: $R=mg\cos\alpha$ , down-slope weight $mg\sin\alpha$ .

1. Sliding down a rough plane

A particle sliding down a rough plane

Perpendicular:

R=mg\cos\alpha

. Along (down positive):

mg\sin\alpha-\mu R=ma

Resolving weight on a slope

$\text{angle between }W\text{ and the normal}=\alpha$
Tilting the plane by $\alpha$ tilts its normal by $\alpha$ from vertical.
$\perp:\ mg\cos\alpha,\qquad\parallel:\ mg\sin\alpha$
Adjacent and opposite components.
$a=g(\sin\alpha-\mu\cos\alpha)$
Sliding down: $mg\sin\alpha-\mu mg\cos\alpha=ma$ ; $m$ cancels.

Worked example3 kg down a $25^\circ$ slope, $\mu=0.3$

Acceleration.

Goal: $a$ .
Strategy: Resolve; friction up the slope.

$R=29.4\cos25^\circ=26.6\text{ N}$
Perpendicular balance.
$F=0.3\times26.6=7.99\text{ N}$
Sliding.
$29.4\sin25^\circ-7.99=3a$
Along: $12.42-7.99$ .
$a=1.48\text{ m s}^{-2}$
Divide.

Answer

1.48

m s

^{-2}

✓ Check

9.8(\sin25^\circ-0.3\cos25^\circ)=1.48

✓.

Worked examplePulled up the same slope by 30 N

Force parallel to the slope.

Goal: $a$ .
Strategy: Friction now acts down the slope.

$30-12.42-7.99=3a$
Both weight component and friction oppose.
$a=3.20\text{ m s}^{-2}$
$9.59/3$ .

Answer

3.20

m s

^{-2}

up the slope.

✓ Check

R

unchanged since the pull is parallel ✓.

Common misconceptionFriction direction Friction opposes the motion (or attempted motion), so it switches when the direction does.

Pause and think

Angle at which a block is about to slip if $\mu=0.5$ ?

Summary

Resolve perpendicular for $R$ .
$F=\mu R$ when moving.
Friction opposes motion.

Rough: Friction acts.
Coefficient of friction: $\mu$ .

HistoryDa Vinci's friction notes. Leonardo da Vinci, around 1500, found by experiment that friction depends on the weight of an object but not on the area in contact. His notebooks weren't published, so Guillaume Amontons rediscovered the laws in 1699.

Fun factWhy ice is slippery. Ice has a very low coefficient of friction (about

0.02

0.1

) because a thin, water-like layer forms on its surface. Scientists still debate exactly how that layer forms, more than

150

years after Michael Faraday first studied it.

Keep reading for free

Create a free account to read the rest of this lesson: key formulae, worked examples, explore the graph, common mistakes, exam tip, guided practice and more. Your answers are checked and your progress is saved.

Create free account Sign in